Question

Question #b4f98

Answer 1

`M_r=55.25`

Explanation:

I'll refer to the acid as `HA`.

Start with the equation:

`HA+NaOHrarrNaA+H_2O`

This tells us that 1 mole of `NaOH` will neutralise 1 mole `HA`.

We can find the number of moles of `NaOH` if we can find its volume.

Density `rho` is given by:

`rho=m/v`

`:.v=m/rho=32.68/0.9705=33.673"ml"`

Since `c=n/v` then we can find the number of moles of `NaOHrArr`

`n_(NaOH)=cxxv=1xx33.673/1000=3.3673xx10^(-2)`

Because the acid and alkali react in a 1:1 ratio we can say that the number of moles of `HA` must be the same.

`:.n_(HA)=3.3673xx10^(-2)`

Now we need to find the mass of `HA` used.

I will use the %w/v value of 62.02 which @Stefan found in another version of the question.

`:.m_(HA)/V_(soln)=0.6202`

You stated that 3ml of solution was used so:

`m_(HA)=0.6202xx3=1.8606"g"`

`:.3.3673xx10^(-2)"mol"` weighs `1.8606"g"`

So 1 mole weighs:

`(1.8606)/(3.3673xx10^(-2))=55.25"g"`

`M_r=55.25`